# How to get other solutions to this Pell like equation $x^2-3y^2=4$?

I have a solution for the equation $$x^2-3y^2=4\tag{1}$$ i.e. $(2,0)$ which is quite trivial, I also found solution for the equation $$x^2-3y^2=1\tag{2}$$ i.e. $(1,0)$, similarly trivial. I know the identity related with Brahmagupta: $$x_1-Ny_1^2=k_1\tag{3}$$ $$x_2-Ny_2^2=k_2\tag{4}$$ $$k_1k_2=(x_1-N_1y_1^2)(x_2-N_2y_2^2)=(x_1x_2-Ny_1y_2)^2-N(x_1y_2-x_2y_1)^2\tag{5}$$ So given a solution for $(1)$ as $(x_1, y_1)$, and for $(2)$ as $(1,0)$, we do not obtain any new solutions for $(1)$ [$k_1=4,k_2=1$].

Moreover with some such diophantine equations as $x^2-10y^2=9$ we cannot generate all solutions with a single solutions (MathWorld page illustrates how we have 3 seeds for different families of solutions)

So how can I get all positive solutions of $(1)$ and in general how to do so. Finding a solution may be done by the Chakravala method but what later?

• do you have a sript about the theory which you can use here? – Dr. Sonnhard Graubner Jun 9 '17 at 17:23
• @Dr.SonnhardGraubner The identity is from Wikipedia and the Chakravala method was taught to us at school but you can also find it in Wikipedia. Both on this page – RE60K Jun 9 '17 at 17:25
• – lhf Jun 9 '17 at 17:27

The equation $x^2-3y^2=4$ is best studied in $\mathbb Z[\sqrt 3]$ using the norm $N(x+y\sqrt 3)=x^2-3y^2$.
If $N(\alpha)=4$ and $N(\beta)=1$, then $N(\alpha\beta^n)=4$ and so $\alpha\beta^n$ gives you several solutions, as long as $\beta\ne1$. In this case, we can take $\beta = 2+1\sqrt3$. You get the linear recurrence $$x_{n+1} = 2x_{n} + 3y_n, \qquad x_0 = 2$$ $$y_{n+1} = \hphantom{2}x_{n} + 2y_n, \qquad y_0 = 0$$
You also get solutions if you can find $\gamma$ with $N(\gamma)=-1$. Then $\alpha\gamma^{2n}$ gives you several solutions. But there is no such $\gamma$ because $x^2 \equiv -1 \bmod 3$ has no solutions.
Beginning with the solutions $(2,0)$ and $(4,2)$ for $x^2 - 3 y^2 = 4,$ all the rest of them come from the linear recurrences $$x_{n+2} = 4 x_{n+1} - x_n,$$ $$y_{n+2} = 4 y_{n+1} - y_n.$$
jagy@phobeusjunior:~$./Pell_Target_Fundamental Automorphism matrix: 2 3 1 2 Automorphism backwards: 2 -3 -1 2 2^2 - 3 1^2 = 1 u^2 - 3 v^2 = 4 Fri Jun 9 13:36:39 PDT 2017 u: 4 v: 2 ratio: 2 SEED BACK ONE STEP 2 , 0 u: 14 v: 8 ratio: 1.75 u: 52 v: 30 ratio: 1.73333 u: 194 v: 112 ratio: 1.73214 u: 724 v: 418 ratio: 1.73206 u: 2702 v: 1560 ratio: 1.73205 u: 10084 v: 5822 ratio: 1.73205 u: 37634 v: 21728 ratio: 1.73205 u: 140452 v: 81090 ratio: 1.73205 u: 524174 v: 302632 ratio: 1.73205 u: 1956244 v: 1129438 ratio: 1.73205 u: 7300802 v: 4215120 ratio: 1.73205 Fri Jun 9 13:36:59 PDT 2017 u^2 - 3 v^2 = 4 jagy@phobeusjunior:~$